Exercise 15.1 - Swiflearn

NCERT Solutions for Class 9th Maths Chapter 15

Probability

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Exercise 15.1

Question 1:

In a cricket match, a batswoman hits a boundary 6 times out of 30 balls she

plays. Find the probability that she did not hit the boundary.

Solution: Consider the event of batswoman not hitting the ball as E. Here, the total number of trials is

30. The batswoman has hit 6 boundaries in her innings.

Therefore,

Number of times the boundary is not hit by her is,

30 – 6 = 24

Now,

[No. of balls she didn't hit to the boundary]P(E) =

[Total no. of balls played by her]

24 4P(E) =

30 5=

So, the required probability is 4

5.

Question 2:

1500 families with 2 children were selected randomly and the following

data were recorded:

Number of girl in a

family 2 1 0

Number of families 475 814 211

Compute the probability of a family, chosen at random, having

(i) 2 girls

(ii) 1 girl

(iii) No girl

Also check whether the sum of these probabilities is 1.

Solution: (i)

Here, total number of the families = 1500

The number of families with 2 girls = 475,

Let the event of selecting the family with 2 girls be X, so,



Probability


No. of families with 2 girlsP(X) =

Total no. of families

475 19P(X) =

1500 60=

Hence, the probability of family having 2 girls is 19

60.

(ii)

The number of families having 1 girl are 814. Consider the event of selecting the family with

1 girl as Y, so,

No. of families with 1 girlP(Y) =


814 407P(Y) =

1500 750=

Hence, the probability of family having 1 girl is 407

750.

(iii)

The number of families having no girl is 211. Consider the event of selecting a family with

no girl as a child be Z, so,

No. of families with no girlP(Z) =


211P(Z) =

1500

Hence, the probability of family having no girls is 211

1500.

Question 3:

[Refer to Example 5, Section 14.4, and Chapter 14.]

Find the probability that a student of the class was born in August.

Solution: Refer to example 5,

40 students from a particular section of Class IX were asked about their birth months and the

following graph was prepared from the obtained data:



Probability


The total number of students = 40.

From the graph, the number of students born in August is 6.

P (student born in august 6 3

40 20=

Hence, the probability that the student born in august is 3

20.

Question 4:

Three coins are tossed simultaneously 200 times with the following

frequencies of different outcomes:

Outcome 3 heads 2 heads 1 heads No heads

Frequency 23 72 77 28

If the three coins are simultaneously tossed again, compute the probability

of 2 head coming up.

Solution: The total number of tosses = 200

Number of getting 2 heads = 72

The no. of times 2 heads have shown up.P(E) =

Total no. of trial

72 9P(E) =

200 25=

Hence, the probability of getting 2 heads is 9

25 .

Question 5:



Probability


An organization selected 2400 families at random and surveyed them to

determine a relationship between income level and the number of vehicles

in the family. The information gathered is listed in the table below:

Suppose a family is chosen. Find the probability that the family chosen is

(i) Earning Rs. 10000−13000 per month and owning exactly 2 vehicles.

(ii) Earning Rs. 16000 or more per month and owning exactly 1 vehicle.

(iii) Earning less than Rs. 7000 per month and does not own any vehicle.

(iv) Earning Rs. 13000−16000 per month and owning more than 2

vehicles.

(v) Owning not more than 1 vehicle.

Solution: Total number of families = 2400.

(i)

Number of families who has earning Rs. 10000 −13000 per month and owning exactly 2

vehicles is 29.

29Probability

2400=

Hence, the probability of families earning 10,000 – 13000 per month and owning exactly 2

vehicles is 29

2400.

(ii)

Number of families who has earning Rs. 16000 or more per month and owning exactly 1

vehicle = 579.

579 193Probability

2400 800= =

Hence, the probability of families earning Rs. 16000 or more per month and owning exactly 1

vehicle is 193

800.



Probability


(iii)

Number of families who has earning less than Rs. 7000 per month and does not own any

vehicle = 10.

10 1Probability

2400 240= =

Hence, the probability of families earning less than Rs. 7000 per month and does not own any

vehicle is 1

240.

(iv)

Number of families who has earning Rs. 13000−16000 per month and owning more than 2

vehicles = 25.

25 1Probability

2400 96= =

Hence, the probability of families earning Rs. 13000−16000 per month and owning more

than 2 vehicles is 1

96.

(v)

Number of families who has owning not more than 1 vehicle.

10 0 1 2 1 160 305 535 469 579 2062= + + + + + + + + + =

2062 1031Probability

2400 1200= =

Probability = 2062

2400=

1031

1200

Hence, the probability of families who has owning not more than 1vehicle is 1031

1200.

Question 6:

A teacher wanted to analyze the performance of two sections of students in

a mathematics test of 100 marks. Looking at their performances, she found

that a few students got under 20 marks and a few got 70 marks or above.

So she decided to group them into intervals of varying sizes as follows:

0−20, 20−30… 60−70, 70−100. Then she formed the following table: [Refer

to table 14.7, chapter 14.]

Marks Number of students 0 − 20

20 − 30

30 − 40

40 − 50

50 − 60

7

10

10

20

20



Probability


60 − 70

70 − above 15

8

Total 90

(i) Find the probability that a student obtained less than 20% in the

mathematics test.

(ii) Find the probability that a student obtained marks 60 or above.

Solution: (i)

Total number of students = 90

The students who scored less than 20% in math’s = 7

Probability = 7

90

(ii)

The students who scored more than 60 or equal to 60 = 23,

Total number of students are 90.

Probability = 23

90

Question 7:

To know the opinion of the students about the subject statistics, a survey of

200 students was conducted. The data is recorded in the following table.

Opinion Number of students

Like

Dislike

135

65

Find the probability that a student chosen at random

(i) Likes statistics

(ii) Does not like it.



Probability


Solution:

The total number of students are 200.

(i)

The students who like statistics = 135

Probability (students who like statistics)135 27

200 40= = .

(ii)

The students who dislike statistics = 65.

Probability (students who dislike statistics) 65 13

200 40= =

Question 8:

Refer to Q.2, Exercise 14.2.

The distance (in km) of 40 engineers from their residence to their place of

work were found as follows.

5 3 10 20 25 11 13 7 12 31

19 10 12 17 18 11 32 17 16 2

7 9 7 8 3 5 12 15 18 3

12 14 2 9 6 15 15 7 6 12

What is the empirical probability that an engineer lives:

(i) Less than 7 km from her place of work?

(ii) More than or equal to 7 km from her place of work?

(iii) Within 𝟏

𝟐 km from her place of work?

Solution: (i)

Total number of engineers = 200

The total number of engineers who have their home in 7 km from work place = 9

Probability = Number of engineers living at a distance of less than 7 km form work place

Total number of engineers

= 9

40.

(ii)



Probability


The total number of engineer who have their home in more than or equal to 7 km from the

work place are 40 - 9 = 31

Probability = 31

40

(iii)

The number of engineers who live within 1

2 km of the work place is 0.

Probability = 0

40 = 0.

Question 9:

Activity: Note the frequency of two – wheeler, three-wheeler and four-

wheelers going past during a time – interval in front of school gate. Find

the probability that any one vehicle out of the total vehicles you have

observed is a two-wheeler.

Solution: The respective frequencies of different types of vehicles after observing in front of the school

gate in time interval 6:30 to 7:30 am are:

Types of vehicles

Frequency

Two-wheelers

Three-wheelers

Four-wheelers

550

250

80

Total number of vehicle,

( ) 550 250 80

880

n S = + +

=

Number of two-wheelers,

( ) 550n E =

Therefore, probability of observing two-wheelers is,

( )

( )550 5

880 8

n E

n s= =

Question 10:

Activity:

Ask all the students in your class to write a 3-digit number. Choose any

student from the room at random. What is the probability that the number

written by him/her is divisible by 3? Remember that a number is divisible

by 3, if the sum of its digits is divisible by 3.



Probability


Solution: A three digit number start from 100 to 999.

Total number of three digit numbers = 999 - 99 = 900.

Multiple of 3 in three digit numbers = {102, 105, ....., 999}

Number of multiples of 3 in three digit number = 900

3 = 300

i.e., n(E) = 300

The probability that the number written by Her

Him is divisible by 3 =

n(E)

n(S) =

300

900 =

𝟏

𝟑.

Hence, the probability that the number written by her / him is divisible by 3 is 𝟏

𝟑.

Question 11: Eleven bags of wheat flour, each marked 5 kg, actually

contained the following weights of flour (in kg):

4.97 5.05 5.08 5.03 5.00 5.06 5.08 4.98 5.04 5.07 5.00

Find the probability that any of these bags chosen at random contains

more than 5 kg of flour.

Solution: The number of bags which contain more than 5 kg of wheat flour is 7.

We have total 11 bags, so the required probability of choosing the bag with more than 5 kg of

wheat flour is

No. of bags with more than 5 kg of flourP(E) =

Total number of bags of flour

7P(E) =

11

Hence, the probability that any of these bags chosen at random contains more than 5 kg of

flour is 7

11

Question 12:

A study was conducted to find out the concentration of Sulphur dioxide in

the air in parts per million (ppm) of a certain city. The frequency

distribution of the data obtained for 30 days is as follows:

Concentration of SO2 (in ppm) Number of days (frequency)

0.00 − 0.04 4

0.04 − 0.08 9



Probability


0.08 − 0.12 9

0.12 − 0.16 2

0.16 − 0.20 4

0.20 − 0.24 2

Total 30

Solution:

The concentration of Sulphur dioxide in the interval 0.12 – 0.16 is on 2 days and the total

number of days is 30.

Probability (concentration of Sulphur dioxide in the interval 0.12 – 0.16) = 2

30=

1

15

Question 13:

The blood groups of 30 students of class VIII are given in the following

frequency distribution table:

Blood group Number of students

A 9

B 6

AB 3

O 12

TOTAL 30

Use this table to determine the probability that a student of this class,

selected at random, has blood group AB.

Solution:

The number of students who have the blood group AB = 3

Total number of students = 30

Required Probability = Number of students having blood group AB

Total number of students

The probability which we require is3

30=

1

10 .

Hence, the required probability is 1

10


Exercise 15.1 - Swiflearn

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